SOLUTION: (3n^2 + 2n^2 + 18n) / 6n^2
I don't understand how that would work becuase it'd be fractions...please help!
Algebra.Com
Question 639067: (3n^2 + 2n^2 + 18n) / 6n^2
I don't understand how that would work becuase it'd be fractions...please help!
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
now, you can set expression equal to zero and calculate
RELATED QUESTIONS
Could you please help me solve this system?
10m-3n=5
5m+2n=-8
I thought I could... (answered by Alan3354)
Factor Completely.
6n^4 + 3n^3 -... (answered by scott8148)
the equation is written to the base 3n-1 instead of base 10 (a,b,c) range from 0 to 3n-2
(answered by richard1234)
I,m having problems simplifying this fraction, can you help please?
(2n^2*3n^3) /... (answered by checkley71)
Simplify the following fractions leaving the answer in index form.
(2n^2 * 3n^3/6n^7)... (answered by stanbon)
Compute the sum (a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + ... +... (answered by ikleyn)
Factor completely: m^3-9m^2n+18n^2m
If anyone could help I would appricate it, I came... (answered by jim_thompson5910,josmiceli)
PLEASE HELP ME!
Proof by math induction (2n-1)^3= n^2(2n^2-1)
I am pretty sure it is... (answered by ikleyn)
prove that the product of any three consecutive numbers is even.
So far I have tried , (answered by josgarithmetic,tenkun,amalm06,ikleyn)